Difference between revisions of "2021 AMC 12B Problems/Problem 16"
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Let <math>g(x)</math> be a polynomial with leading coefficient <math>1,</math> whose three roots are the reciprocals of the three roots of <math>f(x)=x^3+ax^2+bx+c,</math> where <math>1<a<b<c.</math> What is <math>g(1)</math> in terms of <math>a,b,</math> and <math>c?</math> | Let <math>g(x)</math> be a polynomial with leading coefficient <math>1,</math> whose three roots are the reciprocals of the three roots of <math>f(x)=x^3+ax^2+bx+c,</math> where <math>1<a<b<c.</math> What is <math>g(1)</math> in terms of <math>a,b,</math> and <math>c?</math> | ||
Revision as of 20:03, 11 February 2021
Problem
Let 
be a polynomial with leading coefficient 
whose three roots are the reciprocals of the three roots of 
where 
What is 
in terms of 
and 
Solution
Note that 
has the same roots as , if it is multiplied by some monomial so that the leading term is 
they will be equal. We have
![\[f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c\]](http://latex.artofproblemsolving.com/d/4/4/d445d497c73df12477b4f7a987c8c3d844dd9e27.png)
so we can see that
![\[g(x) = \frac{x^3}{c}f(1/x)\]](http://latex.artofproblemsolving.com/7/0/f/70f8e3089ec7eba045d932a0dff5250f7cc3a43c.png)
Therefore
![\[g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}\]](http://latex.artofproblemsolving.com/7/9/9/79989395b4413f6f007ecf21bbe70e43f9bf5984.png)
